Question: Divide the following complex numbers. $ \dfrac{5+3i}{1-i}$
We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${1+i}$ $ \dfrac{5+3i}{1-i} = \dfrac{5+3i}{1-i} \cdot \dfrac{{1+i}}{{1+i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(5+3i) \cdot (1+i)} {(1-i) \cdot (1+i)} = \dfrac{(5+3i) \cdot (1+i)} {1^2 - (-1i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(5+3i) \cdot (1+i)} {(1)^2 - (-1i)^2} = $ $ \dfrac{(5+3i) \cdot (1+i)} {1 + 1} = $ $ \dfrac{(5+3i) \cdot (1+i)} {2} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({5+3i}) \cdot ({1+i})} {2} = $ $ \dfrac{{5} \cdot {1} + {3} \cdot {1 i} + {5} \cdot {1 i} + {3} \cdot {1 i^2}} {2} $ Evaluate each product of two numbers. $ \dfrac{5 + 3i + 5i + 3 i^2} {2} $ Finally, simplify the fraction. $ \dfrac{5 + 3i + 5i - 3} {2} = \dfrac{2 + 8i} {2} = 1+4i $